The points A and B have position vectors i + 2j – 2k and 2i – j + k respectively. The line l has equation r = i − j + 3k + μ(2i – 3j + 4k).

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. M1 Carry out correct method for finding a vector equation for AB
2. A1 Obtain [r =] i + 2j – 2k + λ(i – 3j + 3k) [OE e.g. r = 2i − j + k + λ(−i + 3j – 3k) .]
3. M1 Equate two pairs of components of general points on their AB and l and evaluate λ or μ [e.g. (1+λ = 1+2μ), (2-3λ = -1-3μ), (-2+3λ = 3+4μ) ]
4. A1 Obtain correct answer for λ or μ, e.g. λ = −1, μ = −2 [Correct value from two correct component equations.]
5. A1 Verify that all three equations are not satisfied and the lines fail to intersect (≠ is sufficient justification e.g. 0 ≠ -3). [Conclusion needs to follow correct values. Hybrid versions are possible e.g. using j and k to get one parameter and then i to obtain the other. or e.g. solving two pairs of simultaneous equations and showing that the results are not the same. Alternatives: table for A (λ, μ) and B (λ, μ) components with values.]
6. B1 Find AP for a general point P on l, e.g. -3j + 5k + μ(2i – 3j + 4k) [Or equivalent e.g. PA = −2μi + (3μ+3)j−(4μ+5)k.]
7. M1 Calculate scalar product of their AP and a direction vector for l and equate the result to zero [e.g. 4μ+(9+9μ)+(20+16μ) = 0. M0 if using OP. M0 if using parallel line through A.]
8. A1 Obtain μ = -1
9. A1 Obtain answer -i + 2j – k [Accept coordinates in place of position vector.]
10. Alternative Method for Question 11(b):
11. B1 Find AP for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) [Or equivalent e.g. PA = −2μi +(3μ +3)j−(4μ+5)k .]
12. M1 Use Pythagoras and differentiate with respect to μ to obtain value of μ corresponding to minimum distance. (No need to prove it is a minimum) [d/dμ(4μ² +9(μ+1)² + (4μ+5)²) = 0.]
13. A1 Obtain μ = -1
14. A1 Obtain answer –i + 2j – k [Accept coordinates in place of position vector.]